The graph of the rational function $\frac{p(x)}{q(x)}$ is shown below. If $q(x)$ is quadratic, $p(3)=3$, and $q(2) = 2$, find $p(x) + q(x)$.

[asy]
size(8cm);
import graph;

Label f; 
f.p=fontsize(6); 
//xaxis(-5,5,Ticks(f, 1.0)); 
//yaxis(-5,5,Ticks(f, 1.0));
draw((-5,0)--(5,0));
draw((0,-5)--(0,5));

int i;

for (i = -5; i <= 5; ++i) {
  if (i != 0) {
    draw((i,-0.2)--(i,0.2));
    draw((-0.2,i)--(0.2,i));
   label("$" + string(i) + "$", (i,-0.2), S);
   label("$" + string(i) + "$", (-0.2,i), W);
  }
}

real f(real x) {return x/((x-1)*x);}

draw(graph(f,-5,-3.5), dashed);
draw(graph(f,-3.5,-0.1));
draw(graph(f,0.1,0.7));
draw(graph(f,0.7,0.8), dashed);
draw(graph(f,1.2,1.3), dashed);
draw(graph(f,1.3,3.5));
draw(graph(f,3.5,5), dashed);
filldraw(circle((0,-1),.15),white);
[/asy]
Solution: Since $q(x)$ is quadratic, and we have a horizontal asymptote at $y=0,$ we know that $p(x)$ must be linear.

Since we have a hole at $x=0,$ there must be a factor of $x$ in both $p(x)$ and $q(x).$ Lastly, since there is a vertical asymptote at $x=1,$ the denominator $q(x)$ must have a factor of $x-1.$ Then, $p(x) = ax$ and $q(x) = bx(x-1),$ for some constants $a$ and $b.$ Since $p(3) = 3,$ we have $3a = 3$ and hence $a=1.$ Since $q(2) = 2,$ we have $2b(2-1) = 2$ and hence $b=1.$

So $p(x) = x$ and $q(x) = x(x - 1) = x^2 - x,$ and $p(x) + q(x) = \boxed{x^2}$.